How do you integrate #int x/(sqrt(2x-1))dx# from [1,5]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Narad T. Dec 19, 2016 The answer is #=16/3# Explanation: #x=1/2(2x-1)+1/2# Let's rewrite the function #x/(sqrt(2x-1))=(1/2(2x-1)+1/2)/(sqrt(2x-1))# #=1/2*(2x-1)/(sqrt(2x-1))+1/(2sqrt(2x-1))# #=1/2*sqrt(2x-1)+1/2*1/sqrt(2x-1)# #int_1^5(xdx)/(sqrt(2x-1))=int_1^5 1/2*sqrt(2x-1)dx+int_1^5 1/2*dx/sqrt(2x-1)# #= [ 1/2*(2x-1)^(3/2)/(2*3/2)+1/2*(2x-1)^(1/2)/(2*1/2) ]_1^5 # #= [(2x-1)^(3/2)/6+(2x-1)^(1/2)/2 ]_1^5 # #=(9/2+3/2)-(1/6+1/2)# #=6-2/3=16/3# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 3846 views around the world You can reuse this answer Creative Commons License